| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2002 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | CI from raw data list |
| Difficulty | Standard +0.3 This is a standard S3 confidence intervals and hypothesis testing question requiring routine application of formulas. Part (a) uses standard unbiased estimator formulas, parts (b-c) apply normal distribution tables with given σ, and part (d) is a one-tailed z-test. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks |
|---|---|
| Cooling by subtracting 500 for each observation gives | M1 A1 |
| Mean \(= 500 + \frac{22}{10} = 502.2\) | |
| Variance \(= \frac{1}{9}\left\{288 - \frac{22^2}{10}\right\} = 26.622\) | M1 A1 A1 (5) |
| Answer | Marks |
|---|---|
| Limits are \(502.2 \pm 1.6449 \times 5.0\) | M1 |
| \((493.98, 510.42)\) [accept \((494, 510)\)] | A1 (2) |
| Answer | Marks |
|---|---|
| 95% confidence limits are | M1 A1ft |
| \(502.2 \pm 1.96 \times \frac{5.0}{\sqrt{10}}\) | B1 (for 1.96) |
| \((499, 505)\) | A1 A1 (5) |
| Answer | Marks |
|---|---|
| \(H_0: \mu = 500\) | B1 (both) |
| \(H_1: \mu > 500\) | |
| \(\alpha = 0.05 \Rightarrow \text{CR: } z > 2.3263\) | B1 |
| \(z = \frac{503.9 - 500}{5.0/\sqrt{15}} = 1.47\) | M1 A1 |
| \(1.47\) is not in the critical region \(\Rightarrow\) no evidence to reject \(H_0\); no evidence to suggest mean is greater than \(500g\) | A1 ft (5) |
| Total: 17 marks |
## Part (a)
| Cooling by subtracting 500 for each observation gives | M1 A1 | |
| Mean $= 500 + \frac{22}{10} = 502.2$ | | |
| Variance $= \frac{1}{9}\left\{288 - \frac{22^2}{10}\right\} = 26.622$ | M1 A1 A1 (5) | |
## Part (b)
| Limits are $502.2 \pm 1.6449 \times 5.0$ | M1 | |
| $(493.98, 510.42)$ [accept $(494, 510)$] | A1 (2) | |
## Part (c)
| 95% confidence limits are | M1 A1ft | |
| $502.2 \pm 1.96 \times \frac{5.0}{\sqrt{10}}$ | B1 (for 1.96) | |
| $(499, 505)$ | A1 A1 (5) | |
## Part (d)
| $H_0: \mu = 500$ | B1 (both) | |
| $H_1: \mu > 500$ | | |
| $\alpha = 0.05 \Rightarrow \text{CR: } z > 2.3263$ | B1 | |
| $z = \frac{503.9 - 500}{5.0/\sqrt{15}} = 1.47$ | M1 A1 | |
| $1.47$ is not in the critical region $\Rightarrow$ no evidence to reject $H_0$; no evidence to suggest mean is greater than $500g$ | A1 ft (5) | |
| **Total: 17 marks** | | |The weights of tubs of margarine are known to be normally distributed. A random sample of 10 tubs of margarine were weighed, to the nearest gram, and the results were as follows.
498 502 500 496 509 504 511 497 506 499
\begin{enumerate}[label=(\alph*)]
\item Find unbiased estimates of the mean and the variance of the population from which this sample was taken. [5]
\end{enumerate}
Given that the population standard deviation is 5.0 g,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item estimate limits, to 2 decimal places, between which 90\% of the weights of the tubs lie, [2]
\item find a 95\% confidence interval for the mean weight of the tubs. [5]
\end{enumerate}
A second random sample of 15 tubs was found to have a mean weight of 501.9 g.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Stating your hypotheses clearly and using a 1\% level of significance, test whether or not the mean weight of these tubs is greater than 500 g. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2002 Q7 [17]}}